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Class 11 Physics Chapter 7 Gravitation — Formulas & Key Points

Chapter 7 Gravitation in NCERT Class 11 Physics introduces the fundamental force that governs planetary orbits, satellite motion, and the weight of objects. This formula sheet presents every critical equation, constant, definition, and law from the chapter in a structured table format for quick revision. Use it to prepare for board exams, NEET, and JEE Main where gravitation problems carry significant weightage each year.

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Key takeaways

  • ✓Newton's law of universal gravitation: F = G m₁m₂/r², where G = 6.67×10⁻¹¹ N·m²/kg² is the universal gravitational constant.
  • ✓Acceleration due to gravity g = GM/R² at Earth's surface; varies with height and depth but is approximately 9.8 m/s² at sea level.
  • ✓Gravitational potential energy U = –GMm/r; gravitational potential V = –GM/r; both are zero at infinite separation.
  • ✓Orbital velocity v₀ = √(GM/r) and escape velocity vₑ = √(2GM/R) = √2 times orbital velocity at the surface.
  • ✓Kepler's three laws govern planetary motion: law of orbits (ellipse), law of areas (equal areas in equal times), law of periods (T² ∝ a³).
  • ✓Weight W = mg is a force measured in Newtons; mass m is constant everywhere but weight changes with gravitational field strength.
  • ✓Satellites in geostationary orbit have period 24 hours, orbital radius ~42,000 km from Earth's center, and zero inclination to the equator.

All Formulas & Laws — Quick Reference Table

This table lists every formula you need from Chapter 7 Gravitation. Each row specifies the formula name, the exact mathematical expression, SI units for all terms, and the typical problem context where you apply it. Bookmark this section for the night before your exam. All symbols follow NCERT conventions: M and m denote masses, r is distance between centers, R is radius of the planet or Earth, h is height above surface, d is depth below surface, G is the universal gravitational constant, g is acceleration due to gravity, T is time period, a is semi-major axis, and v denotes velocity. Pay close attention to negative signs in potential and potential energy—they indicate attractive forces and bound systems. The table format ensures you can scan and locate any formula in seconds during last-minute revision or while solving numericals.
  • Newton's Law of Universal Gravitation: F = G m₁m₂/r² (units: N). Use whenever two point masses or spherically symmetric bodies attract each other.
  • Acceleration due to gravity at surface: g = GM/R² (units: m/s²). Use to find g for any planet given its mass M and radius R.
  • Weight of an object: W = mg (units: N). Use to convert mass to weight or vice versa in any gravitational field.
  • Gravitational field intensity: I = GM/r² = g (units: N/kg or m/s²). Use to find force per unit mass at distance r from mass M.
  • Gravitational potential energy: U = –GMm/r (units: J). Use for energy calculations; zero at r = ∞, negative for bound systems.
  • Gravitational potential: V = –GM/r (units: J/kg). Use for potential per unit mass; independent of test mass m.
  • Variation of g with height h (h << R): gₕ = g(1 – 2h/R) or gₕ = g(R/(R+h))² for exact value (units: m/s²).
  • Variation of g with depth d: gₐ = g(1 – d/R) (units: m/s²). At Earth's center (d = R), g = 0.
  • Orbital velocity: v₀ = √(GM/r) (units: m/s). Use for satellites or planets in circular orbits at distance r from center.
  • Time period of satellite: T = 2π√(r³/GM) (units: s). Use to find how long one complete orbit takes.
  • Escape velocity: vₑ = √(2GM/R) = √(2gR) (units: m/s). Minimum speed to escape gravitational field starting from surface.
  • Energy of orbiting satellite: Total E = –GMm/(2r) = K.E. + P.E. Kinetic energy K.E. = +GMm/(2r), Potential energy P.E. = –GMm/r.
  • Kepler's third law: T² ∝ a³ or T²/a³ = 4π²/GM (units: s²/m³). For planets orbiting the Sun or satellites orbiting a planet.

Key Terms, Definitions & Conceptual Points

Gravitation is the universal attractive force between any two masses; it is always attractive, never repulsive, and acts along the line joining the centers of the masses. The gravitational constant G = 6.67×10⁻¹¹ N·m²/kg² is a universal constant, the same everywhere in the universe, measured by Cavendish in 1798 using a torsion balance experiment. Acceleration due to gravity g is the acceleration imparted to a freely falling body and depends on the mass and radius of the planet; on Earth g ≈ 9.8 m/s², on Moon g ≈ 1.6 m/s². Weight is the gravitational force exerted by a planet on an object and varies with location; mass is the quantity of matter and remains constant everywhere. Gravitational potential energy is the work done to bring a mass from infinity to a point in a gravitational field; it is zero at infinity and negative at finite distances, reflecting the attractive nature of gravity. Escape velocity is the minimum velocity required to project a body from the surface of a planet so that it escapes to infinity and never returns; it is independent of the mass and direction of the projectile. A satellite is a body that revolves around a planet in a stable orbit under gravitational attraction; geostationary satellites have a 24-hour period and remain fixed above one point on the equator, used for communication and weather monitoring.
  • Gravitational force obeys inverse-square law: doubling distance makes force one-fourth; tripling makes it one-ninth.
  • Gravitational field is a region around a mass where another mass experiences a force; field strength equals g.
  • Binding energy of a satellite is the energy needed to remove it from orbit to infinity; equals |Total Energy|.
  • Orbital velocity decreases with altitude: higher orbits are slower; satellites closer to Earth move faster.
  • Kepler's first law: planets move in elliptical orbits with the Sun at one focus.
  • Kepler's second law: a line joining planet and Sun sweeps equal areas in equal times (angular momentum conservation).
  • Kepler's third law relates period T and semi-major axis a: larger orbits take longer to complete.
  • Weightlessness in orbit occurs because satellite and occupants fall together; not because gravity is absent.

Important Constants & Standard Values

Memorise these constants and values exactly as they appear in NCERT and CBSE mark schemes because using incorrect values will lose you marks even if your method is perfect. The universal gravitational constant G = 6.67×10⁻¹¹ N·m²/kg² or 6.67×10⁻¹¹ m³ kg⁻¹ s⁻² in SI base units. Acceleration due to gravity at Earth's surface g = 9.8 m/s² (use 10 m/s² only if the question explicitly says so or for rough estimates). Mass of Earth M = 6×10²⁴ kg, radius of Earth R = 6.4×10⁶ m or 6400 km. Mass of Sun = 2×10³⁰ kg. Mass of Moon = 7.4×10²² kg, radius of Moon = 1.74×10⁶ m. Mean Earth-Moon distance = 3.84×10⁸ m. Mean Earth-Sun distance (1 Astronomical Unit) = 1.5×10¹¹ m. Radius of geostationary orbit from Earth's center ≈ 42,000 km or 4.2×10⁷ m. Always write units in your final answer; omitting units costs marks in CBSE board exams and competitive exams alike.
  • Universal gravitational constant: G = 6.67×10⁻¹¹ N·m²/kg² (exact value to three significant figures).
  • Standard gravity: g = 9.8 m/s² on Earth at sea level; g = 1.6 m/s² on Moon; g = 3.7 m/s² on Mars.
  • Earth's mass: M⊕ = 6×10²⁴ kg; Earth's radius: R⊕ = 6.4×10⁶ m.
  • Sun's mass: M☉ = 2×10³⁰ kg (about 333,000 times Earth's mass).
  • 1 Astronomical Unit (AU): 1.5×10¹¹ m (average Earth-Sun distance).
  • Geostationary orbit radius: ~4.2×10⁷ m from Earth's center (~36,000 km above surface).
  • Speed of light c = 3×10⁸ m/s (useful for some astrophysics problems).

Memory Tricks, Mnemonics & Quick Recall Tips

Use these mnemonics and memory aids to remember formulas under exam pressure. For Newton's law F = Gm₁m₂/r², think 'Greater Masses, Smaller Radius → Stronger Force' to recall the direct and inverse relationships. To remember that gravitational potential and potential energy are negative, think 'Gravity Pulls Inward, Energy Goes Negative' — you have to do work to pull things apart, so bound states have negative energy. Orbital velocity v₀ = √(GM/r) and escape velocity vₑ = √(2GM/R) differ by a factor of √2; remember 'Escape is √2 times Orbit'. For variation with height gₕ = g(1 – 2h/R), the coefficient 2 comes from differentiating 1/r²; just recall 'twice the height fraction'. Kepler's third law T² ∝ a³ can be remembered as 'Time-Squared, Distance-Cubed'. Weight W and mass m: 'Weight Wavers, Mass is Constant' — weight changes with location, mass does not. For the sign of work done against gravity, remember 'Against Gravity, Work Positive; With Gravity, Work Negative'. These mnemonics are especially helpful in the last 15 minutes before the exam when you want to do a final mental check of all formulas.
  • G-MEN: G for Gravitational constant, M for Mass, E for Earth, N for Newton — helps recall G and Newton's law.
  • VEO: Velocity-Escape-Orbital → vₑ = √2 v₀ at the same point (usually surface).
  • Negative potential: 'Zero at infinity, negative when near' — potential energy is most negative at closest approach.
  • Kepler mnemonic: 'Orbits are Ellipses, Areas are Equal, Periods are Proportional' for the three laws in order.
  • Height reduces g by '2h/R', depth reduces g by 'd/R' — notice the factor of 2 only for height (from calculus).
  • Total energy of orbit is half the potential energy: E = U/2 = –GMm/(2r), kinetic energy K = –E.
  • Escape velocity does not depend on the mass of the escaping object or its direction — only on planet's M and R.

Common Mistakes, Sign Errors & Unit Traps

Students repeatedly lose marks on gravitation numericals by making the same avoidable errors every year. First, radius versus diameter confusion: always use radius r or R in formulas, not diameter; if the question gives diameter, halve it immediately and write r = D/2 in your working. Second, forgetting the negative sign in gravitational potential energy U = –GMm/r and gravitational potential V = –GM/r; this negative sign is not optional, it indicates bound systems and attractive forces. Third, using g when you should use G or vice versa: g (lowercase, ~9.8 m/s²) is acceleration due to gravity specific to a location, while G (uppercase, 6.67×10⁻¹¹) is the universal gravitational constant. Fourth, unit inconsistency: convert all distances to meters, masses to kilograms, time to seconds before substituting into formulas. Fifth, confusing orbital velocity and escape velocity: orbital velocity keeps a satellite in circular orbit, escape velocity sends it to infinity; escape is always larger. Sixth, writing weight in kilograms instead of Newtons: weight is a force, so its unit is Newton (N), not kilogram. Seventh, in variation of g with height, using h when you should use (R+h) in the exact formula gₕ = GM/(R+h)²; the approximate formula gₕ = g(1 – 2h/R) is valid only when h << R.
  • Always square the entire denominator (r² or (R+h)²), not just R or r separately.
  • Negative sign in U and V: write it in the formula from the start, do not add or drop it arbitrarily.
  • Check units in the final answer: force in N, energy in J, velocity in m/s, distance in m, mass in kg.
  • Gravitational force is always attractive; there is no repulsive gravity, so F is always positive in magnitude.
  • When calculating g on another planet, use that planet's M and R, not Earth's values.
  • In Kepler's third law T²/a³ = constant, use consistent units: if T is in years and a in AU, constant = 1; if SI units, constant = 4π²/GM.
  • Escape velocity formula uses radius from center to surface R, not the altitude h above surface.
  • Do not cancel G with g; they are completely different constants with different units and meanings.

Worked Example 1 — Newton's Law & Weight Calculation

Problem: An object of mass 10 kg is placed on the surface of a planet with mass 5×10²⁴ kg and radius 4×10⁶ m. Calculate (i) the acceleration due to gravity on the planet's surface, (ii) the weight of the object, and (iii) the gravitational force between the planet and the object. Use G = 6.67×10⁻¹¹ N·m²/kg². Solution walkthrough with every algebraic step shown clearly. Part (i): g = GM/R² = (6.67×10⁻¹¹ × 5×10²⁴)/(4×10⁶)² = (3.335×10¹⁴)/(1.6×10¹³) = 20.84 m/s² ≈ 21 m/s². Part (ii): W = mg = 10 × 20.84 = 208.4 N ≈ 208 N. Part (iii): Gravitational force F = GMm/r² = (6.67×10⁻¹¹ × 5×10²⁴ × 10)/(4×10⁶)² = (3.335×10¹⁵)/(1.6×10¹³) = 208.4 N. Notice that weight W and gravitational force F are identical because weight is simply the gravitational force exerted by the planet on the object. This equivalence is a key conceptual point often tested in CBSE exams.

Worked Example 2 — Orbital & Escape Velocity

Problem: A satellite is orbiting Earth at a height of 400 km above the surface. Earth's radius R = 6400 km = 6.4×10⁶ m, mass M = 6×10²⁴ kg. Calculate (i) orbital velocity, (ii) time period, (iii) escape velocity from that height. Use G = 6.67×10⁻¹¹ N·m²/kg². Solution: Total distance from Earth's center r = R + h = 6.4×10⁶ + 0.4×10⁶ = 6.8×10⁶ m. Part (i): Orbital velocity v₀ = √(GM/r) = √[(6.67×10⁻¹¹ × 6×10²⁴)/(6.8×10⁶)] = √[(4.002×10¹⁴)/(6.8×10⁶)] = √(5.885×10⁷) ≈ 7670 m/s ≈ 7.67 km/s. Part (ii): Time period T = 2πr/v₀ = (2 × 3.14 × 6.8×10⁶)/7670 = (4.27×10⁷)/7670 ≈ 5567 seconds ≈ 92.8 minutes ≈ 1.55 hours (typical for low Earth orbit). Part (iii): Escape velocity from height h: vₑ = √(2GM/r) = √2 × v₀ = 1.414 × 7670 ≈ 10,850 m/s ≈ 10.85 km/s. Notice escape velocity from 400 km altitude is slightly less than from surface (11.2 km/s) because the satellite is already higher in the gravitational well.

Worked Example 3 — Kepler's Third Law Application

Problem: The mean distance of Mars from the Sun is 1.52 times the mean distance of Earth from the Sun. If Earth's orbital period is 1 year, calculate the orbital period of Mars using Kepler's third law. Solution: Kepler's third law states T² ∝ a³, or T₁²/T₂² = a₁³/a₂³ for two planets orbiting the same star. Let subscript 1 denote Earth, subscript 2 denote Mars. Given: a₂ = 1.52 a₁, T₁ = 1 year. We need T₂. Write the ratio: (T₁/T₂)² = (a₁/a₂)³ = (a₁/(1.52 a₁))³ = (1/1.52)³ = (0.658)³ ≈ 0.2846. Therefore T₁²/T₂² = 0.2846, so T₂² = T₁²/0.2846 = (1 year)²/0.2846 ≈ 3.514 year². Taking square root: T₂ = √3.514 ≈ 1.875 years ≈ 1.88 years. (The actual Martian year is 1.88 Earth years, confirming Kepler's law precisely.) This problem is a favourite in CBSE board exams because it tests proportional reasoning and does not require knowing G or masses explicitly, only the ratio of distances.

Variation of g — Height, Depth & Rotation Effects

Acceleration due to gravity g is not constant across Earth; it varies with altitude, depth, latitude, and local geology. At height h above Earth's surface, g decreases because distance from Earth's center increases. For small h (h << R), use the approximate formula gₕ = g(1 – 2h/R); for exact calculations use gₕ = GM/(R+h)² = g[R/(R+h)]². At Mount Everest summit (h ≈ 8.8 km), g is about 0.27% less than at sea level, a small but measurable difference. As you go below Earth's surface to depth d, g decreases linearly: gₐ = g(1 – d/R), reaching zero at Earth's center (d = R) because the mass 'above' you cancels out symmetrically. Earth's rotation causes an apparent reduction in g at the equator due to centrifugal effect; g is maximum at the poles (~9.83 m/s²) and minimum at the equator (~9.78 m/s²), a difference of about 0.5%. The value of g also varies slightly due to density variations in Earth's crust — regions with dense minerals or rock have slightly higher g, used in geological surveys and oil exploration. For CBSE numericals, you will mostly use the height and depth formulas; know both approximate and exact forms and choose based on what the question provides.
  • Height formula (exact): gₕ = g × [R/(R+h)]²; when h << R, approximate as gₕ ≈ g(1 – 2h/R).
  • Depth formula: gₐ = g(1 – d/R); linear decrease, not inverse-square, because enclosed mass decreases.
  • At the center of Earth (d = R): g = 0, so objects would be weightless there (but not massless).
  • Gravity decreases by ~0.03% per kilometer of altitude gain near Earth's surface.
  • Rotation effect: effective g = g – ω²R cosθ, where θ is latitude; maximum reduction at equator (θ=0°).
  • Poles have higher g than equator by ~0.5% due to both rotation and Earth's oblate shape.
  • Weightlessness in space is not due to zero gravity but due to free-fall (gravity is still present but not felt).

Satellites, Orbits & Geostationary Conditions

A satellite is any object that revolves around a planet under gravitational attraction, including natural satellites like the Moon and artificial satellites launched by humans for communication, GPS, weather monitoring, and scientific research. For a stable circular orbit, gravitational force provides the necessary centripetal force: GMm/r² = mv²/r, which simplifies to v = √(GM/r), the orbital velocity formula. The total mechanical energy of a satellite in orbit is E = –GMm/(2r), which is negative, indicating a bound system; kinetic energy K.E. = +GMm/(2r) and potential energy P.E. = –GMm/r, so E = K.E. + P.E. = GMm/(2r) – GMm/r = –GMm/(2r). A geostationary satellite appears stationary above a fixed point on Earth's equator because its orbital period matches Earth's rotation period (24 hours). The conditions for a geostationary orbit are: period T = 24 hours = 86400 seconds, orbit must be circular, orbit must lie in the equatorial plane (zero inclination), and orbital radius from Earth's center is approximately 42,000 km (altitude ~36,000 km above surface). These satellites are crucial for TV broadcasting, telecommunications, and meteorology. Polar satellites, by contrast, orbit over the poles at much lower altitudes (~800 km), completing many orbits per day and scanning the entire Earth as it rotates beneath them; used for mapping, surveillance, and Earth observation.
  • Orbital velocity decreases with altitude: v₀ = √(GM/r), so higher orbits are slower.
  • Time period increases with altitude: T = 2π√(r³/GM); geostationary orbit has T = 24 hours.
  • Total energy E = –GMm/(2r) = (1/2) × P.E.; satellite's K.E. = –E = GMm/(2r).
  • Binding energy (energy to remove satellite to infinity) = |E| = GMm/(2r).
  • Geostationary orbit radius r ≈ 4.2×10⁷ m from Earth's center, altitude h ≈ 3.6×10⁷ m above surface.
  • Low Earth orbit (LEO): 200-2000 km altitude, period ~90 minutes; used for ISS, imaging satellites.
  • Polar orbit: passes over both poles, allows complete Earth coverage as planet rotates below.
  • Escape velocity from orbit at radius r: vₑ = √(2GM/r) = √2 × v₀; independent of satellite mass.

One-Glance Last-Minute Revision Box

Use this box for final revision 10 minutes before the exam. Core formulas: F = Gm₁m₂/r² (gravitational force), g = GM/R² (surface gravity), W = mg (weight), U = –GMm/r (potential energy), V = –GM/r (potential), v₀ = √(GM/r) (orbital velocity), vₑ = √(2GM/R) = √(2gR) (escape velocity), T² ∝ a³ (Kepler III). Variation: gₕ = g(1–2h/R) for height, gₐ = g(1–d/R) for depth. Energy of orbit: E = –GMm/(2r), K.E. = GMm/(2r), P.E. = –GMm/r. Constants: G = 6.67×10⁻¹¹ N·m²/kg², g = 9.8 m/s² on Earth, Earth mass = 6×10²⁴ kg, Earth radius = 6.4×10⁶ m. Key points: gravity is always attractive; weight changes with location, mass does not; escape velocity is √2 times orbital velocity at same point; geostationary orbit has T = 24 h, r ≈ 42,000 km from center; potential energy is negative for bound systems; at Earth's center g = 0. Common errors: forgetting negative sign in U and V, confusing G with g, using diameter instead of radius, writing weight in kg instead of Newtons. Mnemonics: VEO (Velocity Escape = √2 Orbital), 'Twice height' for gₕ variation, 'Zero at infinity, negative near' for potential. Practice dimensional analysis: check that force has units of Newton (kg·m/s²), energy has Joules (kg·m²/s²), velocity has m/s. This box is your friend when you are sitting outside the exam hall doing last-minute mental revision; read it, close your eyes, and recall each formula one by one.

CBSE Exam Tips & How CBSETUTOR.ai Helps

In CBSE Class 11 Physics board exams and internal assessments, Chapter 7 Gravitation typically carries 6-8 marks across short-answer and long-answer questions, plus 2-3 marks in the multiple-choice section. Numerical problems on orbital velocity, escape velocity, and variation of g are extremely common; derivations of formulas (especially orbital velocity and escape velocity) and statement-proof of Kepler's laws appear regularly in 3-mark and 5-mark questions. Always start numerical solutions by writing 'Given:', then list all known values with units, followed by 'To find:', then the formula, substitution, calculation, and final answer with correct units and significant figures. For derivations, begin from first principles (Newton's law, centripetal force, energy conservation) and show every algebraic step; examiners award step-wise marks, so even if the final answer is wrong, clear working earns partial credit. Gravitation concepts also appear in competitive exams like NEET (especially satellite motion, gravitational potential energy) and JEE Main (Kepler's laws, energy in orbits, variation of g). To master this chapter, solve all NCERT in-text questions, end-of-chapter exercises, and previous years' board questions. CBSETUTOR.ai provides 24×7 AI-powered doubt resolution where you can upload a photo of any gravitation problem — whether numerical, derivation, or conceptual — and receive step-by-step solutions instantly, just like having a personal tutor at home. The platform costs a flat ₹999 per month for all subjects and all classes (6-12), includes a 3-day free trial, and is trusted by thousands of CBSE students across India to clarify tough Physics numericals, practice additional problems, and revise formulas interactively before exams.
  • Practice numerical problems daily: 5 problems per day builds speed and accuracy for board exams.
  • Memorise all formulas in the one-glance box; write them on the first page of your answer sheet as soon as exam starts.
  • Derivations: write assumptions clearly (e.g. 'assuming circular orbit', 'point masses'), show all steps, box the final formula.
  • In MCQs, eliminate options using dimensional analysis and order-of-magnitude estimates before calculating exactly.
  • Revise variation of g, orbital and escape velocity, and Kepler's laws — these are high-weightage topics every year.
  • Use standard values (G, g, M, R of Earth) unless the question provides different ones; write values in scientific notation.
  • For conceptual questions (e.g. 'why is gravitational PE negative?'), give clear physical reasoning, not just formula.
  • CBSETUTOR.ai lets you snap a photo of any problem and get instant, curriculum-aligned step-by-step solutions anytime.

Frequently asked questions

What is the difference between gravitational constant G and acceleration due to gravity g?+
G is the universal gravitational constant (6.67×10⁻¹¹ N·m²/kg²), same everywhere in the universe. g is the acceleration due to gravity at a specific location (e.g. 9.8 m/s² on Earth's surface), calculated using g = GM/R². G is a fundamental constant; g depends on the mass and radius of the planet or body you are on.
Why is gravitational potential energy negative?+
Gravitational potential energy U = –GMm/r is negative because we define U = 0 at infinite separation. As two masses come closer, gravity (an attractive force) does positive work, so the system loses potential energy, making U negative. A negative U indicates a bound system where energy must be supplied to separate the masses to infinity.
How do I remember the formula for escape velocity?+
Escape velocity vₑ = √(2GM/R) or vₑ = √(2gR). Remember that it is exactly √2 times the orbital velocity at the surface. Mnemonic: 'Escape is root-2 Orbit'. Derive it once from energy conservation and the logic will stick: kinetic energy at surface must equal the magnitude of gravitational potential energy to reach infinity with zero velocity.
Does escape velocity depend on the mass of the object being launched?+
No. Escape velocity vₑ = √(2GM/R) depends only on the mass M and radius R of the planet, not on the mass m of the object. A feather and a rocket require the same launch speed (11.2 km/s from Earth) to escape, though the rocket needs vastly more energy because energy = (1/2)mvₑ² does depend on m.
What are the conditions for a geostationary satellite?+
A geostationary satellite must have: (1) orbital period T = 24 hours to match Earth's rotation, (2) circular orbit, (3) orbit in the equatorial plane (zero inclination), and (4) orbital radius ~42,000 km from Earth's center (~36,000 km altitude). It appears fixed above one point on the equator, ideal for communication and broadcasting.
Why does g become zero at the center of the Earth?+
At depth d below the surface, only the mass within radius (R – d) contributes to gravitational pull; the spherical shell of mass above you exerts zero net force (shell theorem). At Earth's center (d = R), there is no mass 'below' you to pull you in any direction, so g = 0. However, pressure and temperature there are extreme.
How does g vary with altitude and is there a simple formula?+
At small height h above Earth's surface (h << R), use the approximate formula gₕ = g(1 – 2h/R). For exact calculations or large h, use gₕ = g[R/(R+h)]². For example, at h = 6400 km (one Earth radius above surface), gₕ = g/4 = 2.45 m/s² using the exact formula.
What is the physical meaning of Kepler's third law T² ∝ a³?+
Kepler's third law states that the square of a planet's orbital period is proportional to the cube of the semi-major axis (average orbital radius). Physically, it means planets farther from the Sun take much longer to orbit: Mars at 1.52 AU takes 1.88 years, while Neptune at 30 AU takes 165 years. The law follows from Newton's gravitation and centripetal force.
Can I use g = 10 m/s² in CBSE board exam numericals?+
Only if the question explicitly says 'take g = 10 m/s²' or if it is a rough estimate problem. Otherwise, use g = 9.8 m/s² for accuracy. If the question gives a specific value (e.g. 9.81 m/s²), use that. Always write the value you are using in the 'Given' section to avoid confusion and show the examiner your method.
How is weight different from mass, and why does weight change on the Moon?+
Mass is the quantity of matter in an object (in kg), constant everywhere. Weight is the gravitational force W = mg (in Newtons), which depends on local g. On Earth, g ≈ 9.8 m/s²; on Moon, g ≈ 1.6 m/s². So a 60 kg person weighs 60×9.8 = 588 N on Earth but only 60×1.6 = 96 N on the Moon — about one-sixth the Earth weight.
What is binding energy of a satellite and how is it calculated?+
Binding energy is the minimum energy required to remove a satellite from its orbit to infinity (where total energy is zero). For a satellite in circular orbit at radius r, total energy E = –GMm/(2r). Binding energy = |E| = GMm/(2r). The deeper in the gravitational well (smaller r), the greater the binding energy needed to escape.
How does CBSETUTOR.ai help with Class 11 Physics Gravitation numericals?+
CBSETUTOR.ai offers 24×7 AI tutor access where you can upload a photo of any gravitation problem — whether orbital velocity, Kepler's law, or derivation — and get instant step-by-step solutions aligned with NCERT and CBSE marking schemes. At ₹999/month for all subjects (classes 6-12) with a 3-day free trial, it is the most affordable way to get personalized doubt clearing anytime, especially before exams when doubts pile up.

Related resources

Important Questions: CBSE Class 11 Physics Chapter 7 GravitationCBSE Class 11 Physics Chapter 7 Gravitation Worksheet with AnswersCBSE Class 11 Physics Chapter 6 System of Particles and Rotational Motion Worksheet with AnswersImportant Questions: CBSE Class 11 Physics Chapter 6 System of Particles and Rotational MotionAI Tutor for Class 11: The Smart Alternative to TuitionAI Tutor for Class 11 Accountancy: Learn Faster with Instant HelpNCERT Solutions for Class 9 Physics Chapter 7: Motion – Complete Solved GuideClass 9 Mathematics Chapter 2 Polynomials — Formulas & Key Points

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Class 11 Physics Tutor in Rushikonda, Visakhapatnam

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